Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{8x}{36x - 27} \times \dfrac{28x - 21}{4x} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ 8x \times (28x - 21) } { (36x - 27) \times 4x } $ $ z = \dfrac {8x \times 7(4x - 3)} {4x \times 9(4x - 3)} $ $ z = \dfrac{56x(4x - 3)}{36x(4x - 3)} $ We can cancel the $4x - 3$ so long as $4x - 3 \neq 0$ Therefore $x \neq \dfrac{3}{4}$ $z = \dfrac{56x \cancel{(4x - 3})}{36x \cancel{(4x - 3)}} = \dfrac{56x}{36x} = \dfrac{14}{9} $